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Proof step 2 from proof div-ls-02 (previous step)

0 < j0, i ≤ j, 0 ≤ i ⊦ i / j0 = j / j0, i / j0 < j / j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
j / j0 < i / j0, 0 < j0, i ≤ j, 0 ≤ i, 0 ≤ j ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)